6

I have JSON that looks like this:

{
  "MobileSiteContents": {
    "au/en": [
      "http://www.url1.com",
      "http://www.url2.com",

    ],
    "cn/zh": [
      "http://www.url2643.com",

    ]
  }
}

I'm trying to deserialize it into an IEnumerable of classes that look like this:

public class MobileSiteContentsContentSectionItem : ContentSectionItem
{
    public string[] Urls { get; set; }
}

public abstract class ContentSectionItem
{
    public string Culture { get; set; }
}

Is that possible?
I realise I will probably need to use a Custom JsonConverter for this, but can't find any examples.

I started writing a method to convert using JObject.Parse but not sure if this is the correct / most efficient route to go down:

public IEnumerable<MobileSiteContentsContentSectionItem> Parse(string json)
{
    var jobject = JObject.Parse(json);

    var result = new List<MobileSiteContentsContentSectionItem>();

    foreach (var item in jobject.Children())
    {
        var culture = item.Path;
        string[] urls = new[] { "" }; //= this is the part I'm having troble with here...

        result.Add(new MobileSiteContentsContentSectionItem { Culture = culture, Urls = urls });
    }

    return result;
}
4

2 回答 2

6

你在正确的轨道上。以下是您需要进行的更正：

  1. 您正在迭代顶级对象的子对象，期望获得实际上位于更下一层对象中的数据。您需要导航到该属性的值MobileSiteContents并遍历其子项。
  2. 当您使用Children()of 时JObject，请使用允许您将它们转换为JProperty对象的重载；这将使提取所需数据变得更加容易。
  3. culture从项目Name中获取JProperty
  4. 要获取urls，请获取项目Value的JProperty并用于ToObject<string[]>()将其转换为字符串数组。

这是更正后的代码：

public IEnumerable<MobileSiteContentsContentSectionItem> Parse(string json)
{
    var jObject = JObject.Parse(json);

    var result = new List<MobileSiteContentsContentSectionItem>();

    foreach (var item in jObject["MobileSiteContents"].Children<JProperty>())
    {
        var culture = item.Name;
        string[] urls = item.Value.ToObject<string[]>();

        result.Add(new MobileSiteContentsContentSectionItem { Culture = culture, Urls = urls });
    }

    return result;
}

如果您喜欢简洁的代码，可以将其简化为“单行”：

public IEnumerable<MobileSiteContentsContentSectionItem> Parse(string json)
{
    return JObject.Parse(json)["MobileSiteContents"]
        .Children<JProperty>()
        .Select(prop => new MobileSiteContentsContentSectionItem
        {
            Culture = prop.Name,
            Urls = prop.Value.ToObject<string[]>()
        })
        .ToList();
}

演示：

class Program
{
    static void Main(string[] args)
    {
        string json = @"
        {
            ""MobileSiteContents"": {
                ""au/en"": [
                    ""http://www.url1.com"",
                    ""http://www.url2.com"",
                ],
                ""cn/zh"": [
                    ""http://www.url2643.com"",
                ]
            }
        }";

        foreach (MobileSiteContentsContentSectionItem item in Parse(json))
        {
            Console.WriteLine(item.Culture);
            foreach (string url in item.Urls)
            {
                Console.WriteLine("  " + url);
            }
        }
    }

    public static IEnumerable<MobileSiteContentsContentSectionItem> Parse(string json)
    {
        return JObject.Parse(json)["MobileSiteContents"]
            .Children<JProperty>()
            .Select(prop => new MobileSiteContentsContentSectionItem()
            {
                Culture = prop.Name,
                Urls = prop.Value.ToObject<string[]>()
            })
            .ToList();
    }

    public class MobileSiteContentsContentSectionItem : ContentSectionItem
    {
        public string[] Urls { get; set; }
    }

    public abstract class ContentSectionItem
    {
        public string Culture { get; set; }
    }
}

输出：

au/en
  http://www.url1.com
  http://www.url2.com
cn/zh
  http://www.url2643.com
于 2013-10-18T03:35:24.500 回答
3

我使用 Json.Net 尝试了这个并且工作正常。

public IEnumerable<MobileSiteContentsContentSectionItem> Parse(string json)
{
     dynamic jobject = Newtonsoft.Json.JsonConvert.DeserializeObject(json);

     var result = new List<MobileSiteContentsContentSectionItem>();

     var urls = new List<string>();
     foreach (var item in jobject.MobileSiteContents)
     {
         var culture = item.Name;
         foreach(var url in item.Value)
            urls.Add(url.Value);

         result.Add(new MobileSiteContentsContentSectionItem { Culture = culture, Urls = urls.ToArray() });
     }

     return result;
}
于 2013-10-17T16:49:53.787 回答